Question

If f(x) = x² – 3x + 4, then find the values of x satisfying the equation f(x) = f(2x + 1).

Answer 1

Given as

f(x) = x² – 3x + 4.

Let us find the x satisfying f (x) = f (2x + 1).

Here, we have,

f(2x + 1) = (2x + 1)² – 3(2x + 1) + 4

= (2x) ² + 2(2x) (1) + 1² – 6x – 3 + 4

= 4x² + 4x + 1 – 6x + 1

= 4x² – 2x + 2

Then, f (x) = f (2x + 1)

x² – 3x + 4 = 4x² – 2x + 2

4x² – 2x + 2 – x² + 3x – 4 = 0
3x² + x – 2 = 0

3x² + 3x – 2x – 2 = 0

3x(x + 1) – 2(x + 1) = 0

(x + 1)(3x – 2) = 0

x + 1 = 0 or 3x – 2 = 0

x = –1 or 3x = 2

x = –1 or 2/3

Thus, the values of x are –1 and 2/3.